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I am a high school student who has just started reading elementary electromagnetism and am a completely beginner in this subject.

I have read in books that EM waves are nothing but sinusoidal oscillations of electric fields (and simultaneous sinusoidal oscillations of the induced magnetic field, which is obvious, as predicted by the Maxwell's law of induction) that propagate through space at the speed of light c.

I also have come to know that an uncharged conducting shell immersed in an external electric field (as predicted by the Gauss' Law of electricity) will not permit the external electric field lines to enter into the shell and act as the so called Faraday cage.

This is where something struck me as absurd:

If I immerse a conducting shell in a sinusoidally varying and propagating external electric field- then, as the field disturbance reaches the shell, it would in no way penetrate inside it to affect the interior of the shell and the field inside the shell will remain null.

But, that means the EM waves cannot enter the shell interior, doesn't it?

Everything goes wrong here: a transparent conducting shell is a mere paradox? Where has my reasoning gone wrong? What is the source of my great misconception?

I could find no satisfactory answer in the internet; An elementary explanation is needed- for as I said earlier, I am a complete novice and want to clear my beginner level misconceptions.

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Very simply: just because something is true in the static case doesn't automatically make it true in the dynamic case.

To reject a static electrical field from inside a shell, it is sufficient for the charges on the surface on the shell to move however slowly they want in order to arrange themselves so as to cancel the field. This is Gauss's Law.

When an electrical field is changing, the electrons need to move around to adapt themselves to the new situation. This results in two things:

  1. A current that flows (which itself becomes a source of EM radiation)
  2. A delay (because of resistance in the shell) - if electrons move quickly, the current feels the resistance of the shell; this builds up a voltage that 'slows them down'

Because of these two things, an electromagnetic wave is not (fully) rejected by a conducting shell - how well it is rejected depends on the frequency of the wave and the material properties (resistivity, but also magnetic properties that lead to skin effect - outside of the scope of your question but important).

The net result is that a conducting shell will reflect (part of) the EM wave - it acts as a scatterer (the currents flowing in the shell generate an EM field). When it is very small compared to the wave length, then the field it experiences is "quasi static" - in other words the electrons have time to move - and the wave hardly notices the shell is there; when it is large compared to the size of the wave, it becomes a better scatterer.

When we talk about optics (very high frequency EM waves) then wavelength is much, much smaller than the object size, and the atomic properties of the material (conductor, dielectric) become important in the understanding of the behavior - but I don't think that's what your question was about.

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  • $\begingroup$ Yes, now I understand that Gauss' law can be only applied in the case where the electric field is static w.r.t time. That was my primary misconception, right ? $\endgroup$ – Sagnik Nov 25 '14 at 7:00
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    $\begingroup$ @Sagy No Gauss's law applies to time-varying fields too. $\endgroup$ – Rob Jeffries Nov 25 '14 at 7:17
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    $\begingroup$ @Floris Although your answer is accepted by the OP, it seems to have reinforced a misconception about Gauss's law. Consider the case of a time-dependent field that is entirely radial. You can make this any frequency you want, it will always be excluded from the interior. Conversely, any tangential field at any frequency is possible (if you can get it in there). $\endgroup$ – Rob Jeffries Nov 25 '14 at 8:21
  • $\begingroup$ @RobJeffries Yes I did wonder about making a comment to that effect. I upvoted your answer in an attempt to show "there is more to this". But I wonder if you wrote your comment backwards. "entirely radial field" will always be excluded? That contradicts (it seems) what you wrote in your answer - that a radial field can make it through. And I'm not sure either case is true for any frequency. $\endgroup$ – Floris Nov 25 '14 at 8:28
  • $\begingroup$ @Floris - read it again. $\endgroup$ – Rob Jeffries Nov 25 '14 at 8:57
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Your misconception is the application of Gauss's law.

There are solutions to Maxwell's equations (including Gauss's law), that permit an electric field inside the shell.

If there are no charges inside the shell, all Gauss's law tells you is that the number of electric field lines entering the shell must equal the number coming out.

An example of a field that satisfies this would be one which is everywhere perpendicular to a radius vector. But in an EM wave, the E-field is perpendicular to the wave motion, so if you fire the wave towards the shell centre, then this condition is automatically satisfied. Some of the wave (see Floris's answer) can make it through and in no way does this violate Gauss's law.

Edit: A further way of thinking about this, is if there are no interior charges (or currents), the only way you can produce an E-field that is tangential to the surface is through Faraday's law via a changing B-field - i.e. time-dependent fields.

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  • $\begingroup$ Well, yes I understand my misconception of Gauss' law. But, then, the technology of electrostatic shielding, doesn't work for time-varying electric fields, right ? $\endgroup$ – Sagnik Nov 25 '14 at 9:33
  • $\begingroup$ @Sagy Gauss's law applies to time-varying fields. Full stop. Are you asking why Maxwell's equations are true? Your misconception is that the field does get through the shell, even if it is attenuated by Ohmic heating in the conductor. This still satisfies Gauss's law because the E-field is not radial - no field lines cross the shell in either direction. Floris's answer simply relates the degree of attenuation to the wave frequency. The charges in the conductor can't arrange themselves to cancel a tangential field inside the shell. $\endgroup$ – Rob Jeffries Nov 25 '14 at 9:33
  • $\begingroup$ Yes yes, I have deleted that comment that's why. I understood my mistake. @Rob Jeffries $\endgroup$ – Sagnik Nov 25 '14 at 9:34
  • $\begingroup$ @Sagy Floris' answer does handle this. Shielding still works because for a good conductor most of the EM radiation is reflected and then what is transmitted into the shell can be mostly dissipated in the shielding. But an isolated conducting shell cannot perfectly exclude time-varying fields. A sheet of aluminium foil is enough though to prevent your mobile phone working - try it at home! $\endgroup$ – Rob Jeffries Nov 25 '14 at 9:39
  • $\begingroup$ Yes, I get it completely now ! @Rob Jeffries $\endgroup$ – Sagnik Nov 25 '14 at 12:42
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That is the point of a Faraday cage.

Although some of the EM field can penetrate inside the shell, under various circumstances, there is an effect where a conductive shell forms a void in a pervasive EM field. To be clear, though, EM fields do not just 'stop' - they are formed of photons, which are reflected at such an interface.

Consider the microwave oven; the metal grid on the window at the front is there to reflect the microwaves and retain them inside the oven; with the other metal walls this forms a Faraday cage to keep the EM in.

The key is the detail of the conductive shell; reflection/absorption is dictated by both conductivity and the size of the holes in the surface. Thus the microwave's screen has holes large enough to permit light through (wavelength < 1 μm) but not microwaves (wavelength > 10 cm).

We can see a similar effect at gamma frequencies, where gamma rays penetrate even plates of metal, because the metal is really composed of a grid of atoms; gamma rays have wavelengths < 1 pm (10-12 m) compared to inter-atomic spacing of Angstroms (10-10 m, or 100s of gamma wavelengths).

The moral of the tale is that the assumptions of the model matter at the extremes; there are no perfectly continuous, perfectly conducting surfaces, so there are no perfect Faraday cages. But still for most applications, when the limitations are understood, the theory works just fine.

Fun things: an EM wave incident on a metal surface can couple with the free electrons in the metal, creating a travelling wave that is trapped at the surface (since the electrons are collectively called a plasma, this is known as a Surface Plasmon), and this wave of electron motion creates its own travelling EM wave - together this is known as a Surface Plasmon Polariton. Theoretically, it might be possible to make perfect lenses, negative refractive indices and indeed invisibility cloaks with such things.

Gauss is just the beginning of EM.

--Edit for Surface Plasmon Resonance:--

For a little more information on surface plasmons, quoting from :

The resonance condition is established when the frequency of incident photons matches the natural frequency of surface electrons oscillating against the restoring force of positive nuclei

Because the coupling between the incoming photons and the plasma at the surface is a resonant coupling, the exact frequency is very sensitive to the conditions at the surface, and this is used for constructing highly sensitive sensors detecting molecular adsorption onto a surface.

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  • $\begingroup$ I only wanted to know whether the Maxwell's equations of electromagnetism allow the EM waves to pass through the shell. I haven't heard about the surface plasmon, but what you said, it sounded fascinating. I am interested to know more about how the trapping occurs, can you explain it further (elementarily) ? $\endgroup$ – Sagnik Nov 25 '14 at 12:39
  • $\begingroup$ I've added a little information on surface plasmon resonance, but if you have more questions, please do post a new question to attract good answers! $\endgroup$ – Phil H Dec 8 '14 at 10:17

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